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Having trouble solving this simple logarithm problem:

$$\log_3(x^3) - 4\log_9(x) - 5\log_{27}(x^{1/2}) = \log_9(4)$$

I’ve been stuck as I when I solve it I get an answer of $2$, when the actual solution seems to be $x=64$?

Whilst editing and trying to explain my working I found the error I did. Solution: $$ \begin{align} \log_3(x^3) - 4\log_9(x) - 5\log_{27}(x^{1/2}) = \log_9(4)& \\ \log_3(x^3) - {\log_3{x^4}}/{\log_3(9)} - {\log_3(x^{1/2})^5}/{\log_3(27)} = {\log_3(4)}/{\log_3(9)}& \\ \log_3(x^3) - {\log_3{x^4}}/{2} - {\log_3(x^{1/2})^5}/{3} = {2\log_3(2)}/{2}& \\ 6\log_3(x^3)/{6} - {3\log_3{x^4}}/{6} - {2\log_3(x^{1/2})^5}/{6} = {\log_3(2)}& \\ 6\log_3(x^3) - {3\log_3{x^4}} - {2\log_3(x^{1/2})^5} = {\log_3(64)}& \\ \log_3(x^3)^6 - {\log_3(x^4)^3} - {\log_3(x^{1/2})^{10}} = {\log_3(64)}& \\ \log_3(x^6)- {\log_3(x^{5})} = {\log_3(64)}& \\ \log_3{x}= {\log_3(64)}& \end{align} $$ And since $\log_c(a) = \log_c(b)$, $a=b$, the solution is

$$x=64$$

Apologies if there is a nicer solution, I tried my best!

Daniele Tampieri
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Harry
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    Thank you, while trying to edit the formatting and write down my working I actually found the error in my working. – Harry Nov 12 '23 at 00:43
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    @bazingachanga consider answering your own question if you've figured it out – Benjamin Wang Nov 12 '23 at 00:55
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    @bazingachanga I agree with B Wang that you should answer your own question. Doing that is encouraged on this site if user figures out the question he or she asked. – coffeemath Nov 12 '23 at 01:13

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By the exponent rule for logarithms, we have \begin{align*} 3\log _3 x - 4\log _9 x - \frac{5}{2}\log _{27} x = \log _9 4 \end{align*} and by change of base rule we have \begin{align*} 3\log _3 x - 4\cdot\frac{\log _3 x}{\log _3 9} - \frac{5}{2}\cdot\frac{\log _3 x}{\log _3 27} = \frac{\log _3 4}{\log _3 9}\\ 3\log _3 x - 4\cdot\frac{\log _3 x}{2} - \frac{5}{2}\cdot\frac{\log _3 x}{3} = \frac{\log _3 4}{2}\text{.} \end{align*} Simplifying yields \begin{align*} \log _3 x &= 3\cdot\log _3 4\\ \log _3 x &= \log _3 (4^3)\\ x &= 4^3=64\text{.} \end{align*}

nomice
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